TITLE: $\langle \Bbb Q,<\rangle$ is an elementary submodel of $ \langle \Bbb R,< \rangle$
QUESTION [7 upvotes]: I am trying to show that $\langle\Bbb Q,<\rangle$ is an elementary submodel of $\langle\Bbb R,<\rangle$.
I first believed that this problem is quite trivial $-$ I thought all I needed to do was show that $\langle\Bbb Q,<\rangle$ and $\langle\Bbb R,<\rangle$ are elementarily equivalent (which follows since both are dense linear orders without endpoint) and then say that $\Bbb Q$ is obviously contained in $\Bbb R$. However, I'm now questioning myself after examining definitions more closely, in particular those related to this post.
In other words, is it not enough to show that the two models are elementarily equivalent, and one a submodel of the other when trying to show one is an elementary submodel of the other?

REPLY [9 votes]: Another way to show the fact is to notice that the theory of $\bf Q$, the theory of dense linear orderings without endpoints, is satisfied by $\bf R$ and eliminates quantifiers.
If a theory $T$ eliminates quantifiers, then it is true that if we have two structures $M\subseteq N$ both satisfying $T$, then $M\preceq N$ (that follows from the fact that quantifier-free formulas are absolute, which is an easy exercise).
Therefore, whenever you have a model $M$ such that $\operatorname{Th}(M)$ eliminates quantifiers, and $M$ is a submodel of $N$, then to show that it is elementary it is enough to show that they're elementarily equivalent.
In general, $M\subseteq N\wedge M\equiv N$ is strictly weaker than $M\preceq N$ (as I've mentioned in the other post with the example $M=(2{\bf Z},+),N=({\bf Z},+)$).